To obtain the Principal Disjunctive Normal Form (PDNF) of the expression $(p \to q) \land (p \leftrightarrow r)$ without using a truth table, we need to proceed through logical manipulations. Here's a step-by-step process:

Step 1: Express the implications and biconditional in terms of basic logical connectives.

• The implication $p \to q$ can be rewritten as: $p \to q \equiv \neg p \lor q$

• The biconditional $p \leftrightarrow r$ can be rewritten as: $p \leftrightarrow r \equiv (p \land r) \lor (\neg p \land \neg r)$

Thus, the original expression becomes: $(\neg p \lor q) \land ((p \land r) \lor (\neg p \land \neg r))$

Step 2: Apply distribution of conjunction over disjunction.

We distribute $(\neg p \lor q)$ over $(p \land r) \lor (\neg p \land \neg r)$, using the distributive law of logic: $A \land (B \lor C) \equiv (A \land B) \lor (A \land C)$

So, $(\neg p \lor q) \land ((p \land r) \lor (\neg p \land \neg r)) \equiv ((\neg p \lor q) \land (p \land r)) \lor ((\neg p \lor q) \land (\neg p \land \neg r))$

Step 3: Simplify each conjunction.

First term: $(\neg p \lor q) \land (p \land r)$

• Distribute $(\neg p \lor q)$ over $(p \land r)$: $(\neg p \lor q) \land (p \land r) \equiv (\neg p \land p \land r) \lor (q \land p \land r)$

• The term $\neg p \land p$ is always false, so it simplifies to: $(q \land p \land r)$

Second term: $(\neg p \lor q) \land (\neg p \land \neg r)$

• Distribute $(\neg p \lor q)$ over $(\neg p \land \neg r)$: $(\neg p \lor q) \land (\neg p \land \neg r) \equiv (\neg p \land \neg p \land \neg r) \lor (q \land \neg p \land \neg r)$

• The term $\neg p \land \neg p \land \neg r$ simplifies to: $\neg p \land \neg r$

• So the whole expression becomes: $(\neg p \land \neg r) \lor (q \land \neg p \land \neg r)$

Step 4: Combine the results.

Now, we combine the two simplified terms from Step 3: $(q \land p \land r) \lor (\neg p \land \neg r) \lor (q \land \neg p \land \neg r)$

Step 5: Further simplifications (if any).

Observe that $(q \land \neg p \land \neg r)$ is already included in $(\neg p \land \neg r)$ because $\neg p \land \neg r$ doesn't depend on $q$. So, the term $q \land \neg p \land \neg r$ is redundant and can be removed.

Thus, the final PDNF is: $(q \land p \land r) \lor (\neg p \land \neg r)$

This is the Principal Disjunctive Normal Form (PDNF) of the expression $(p \to q) \land (p \leftrightarrow r)$.

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Do you have any questions or would you like more details on any step?

Related Questions:

1. What is the difference between PDNF and CNF?
2. How do we simplify logical expressions using Karnaugh maps?
3. Can the logical equivalence $(p \leftrightarrow r)$ be simplified differently?
4. How do we handle implications and biconditionals in logical expressions?
5. What is the role of distributive and associative laws in simplifying Boolean expressions?

Tip:

Always break down logical connectives into simpler forms (like AND, OR, and NOT) when transforming expressions into PDNF or CNF to make distribution and simplification easier.